3.1095 \(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=285 \[ -\frac {2 a^3 \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 d (a+b)}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (7 a^2 C-7 a b B+7 A b^2+5 b^2 C\right )}{21 b^3 d}+\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 C+5 a^2 b B-a b^2 (5 A+3 C)+3 b^3 B\right )}{5 b^4 d}-\frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-21 a^4 C+21 a^3 b B-7 a^2 b^2 (3 A+C)+7 a b^3 B-b^4 (7 A+5 C)\right )}{21 b^5 d}+\frac {2 (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b^2 d}+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 b d} \]

[Out]

2/5*(5*a^2*b*B+3*b^3*B-5*a^3*C-a*b^2*(5*A+3*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(
1/2*d*x+1/2*c),2^(1/2))/b^4/d-2/21*(21*a^3*b*B+7*a*b^3*B-21*a^4*C-7*a^2*b^2*(3*A+C)-b^4*(7*A+5*C))*(cos(1/2*d*
x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/b^5/d-2*a^3*(A*b^2-a*(B*b-C*a))*(co
s(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/b^5/(a+b)/d+2/5*
(B*b-C*a)*cos(d*x+c)^(3/2)*sin(d*x+c)/b^2/d+2/7*C*cos(d*x+c)^(5/2)*sin(d*x+c)/b/d+2/21*(7*A*b^2-7*B*a*b+7*C*a^
2+5*C*b^2)*sin(d*x+c)*cos(d*x+c)^(1/2)/b^3/d

________________________________________________________________________________________

Rubi [A]  time = 1.29, antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {3049, 3059, 2639, 3002, 2641, 2805} \[ -\frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-7 a^2 b^2 (3 A+C)+21 a^3 b B-21 a^4 C+7 a b^3 B-b^4 (7 A+5 C)\right )}{21 b^5 d}+\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 b B-5 a^3 C-a b^2 (5 A+3 C)+3 b^3 B\right )}{5 b^4 d}-\frac {2 a^3 \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 d (a+b)}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (7 a^2 C-7 a b B+7 A b^2+5 b^2 C\right )}{21 b^3 d}+\frac {2 (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b^2 d}+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 b d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x]),x]

[Out]

(2*(5*a^2*b*B + 3*b^3*B - 5*a^3*C - a*b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/(5*b^4*d) - (2*(21*a^3*b*B +
7*a*b^3*B - 21*a^4*C - 7*a^2*b^2*(3*A + C) - b^4*(7*A + 5*C))*EllipticF[(c + d*x)/2, 2])/(21*b^5*d) - (2*a^3*
(A*b^2 - a*(b*B - a*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b^5*(a + b)*d) + (2*(7*A*b^2 - 7*a*b*B + 7
*a^2*C + 5*b^2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*b^3*d) + (2*(b*B - a*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x]
)/(5*b^2*d) + (2*C*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(7*b*d)

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 3002

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3059

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx &=\frac {2 C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 b d}+\frac {2 \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (\frac {5 a C}{2}+\frac {1}{2} b (7 A+5 C) \cos (c+d x)+\frac {7}{2} (b B-a C) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{7 b}\\ &=\frac {2 (b B-a C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 d}+\frac {2 C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 b d}+\frac {4 \int \frac {\sqrt {\cos (c+d x)} \left (\frac {21}{4} a (b B-a C)+\frac {1}{4} b (21 b B+4 a C) \cos (c+d x)+\frac {5}{4} \left (7 A b^2-7 a b B+7 a^2 C+5 b^2 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{35 b^2}\\ &=\frac {2 \left (7 A b^2-7 a b B+7 a^2 C+5 b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 b^3 d}+\frac {2 (b B-a C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 d}+\frac {2 C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 b d}+\frac {8 \int \frac {\frac {5}{8} a \left (7 A b^2-7 a b B+7 a^2 C+5 b^2 C\right )+\frac {1}{8} b \left (35 A b^2+28 a b B-28 a^2 C+25 b^2 C\right ) \cos (c+d x)+\frac {21}{8} \left (5 a^2 b B+3 b^3 B-5 a^3 C-a b^2 (5 A+3 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{105 b^3}\\ &=\frac {2 \left (7 A b^2-7 a b B+7 a^2 C+5 b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 b^3 d}+\frac {2 (b B-a C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 d}+\frac {2 C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 b d}-\frac {8 \int \frac {-\frac {5}{8} a b \left (7 A b^2-7 a b B+7 a^2 C+5 b^2 C\right )+\frac {5}{8} \left (21 a^3 b B+7 a b^3 B-21 a^4 C-7 a^2 b^2 (3 A+C)-b^4 (7 A+5 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{105 b^4}+\frac {\left (5 a^2 b B+3 b^3 B-5 a^3 C-a b^2 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 b^4}\\ &=\frac {2 \left (5 a^2 b B+3 b^3 B-5 a^3 C-a b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d}+\frac {2 \left (7 A b^2-7 a b B+7 a^2 C+5 b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 b^3 d}+\frac {2 (b B-a C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 d}+\frac {2 C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 b d}-\frac {\left (21 a^3 b B+7 a b^3 B-21 a^4 C-7 a^2 b^2 (3 A+C)-b^4 (7 A+5 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 b^5}-\frac {\left (a^3 \left (A b^2-a (b B-a C)\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{b^5}\\ &=\frac {2 \left (5 a^2 b B+3 b^3 B-5 a^3 C-a b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d}-\frac {2 \left (21 a^3 b B+7 a b^3 B-21 a^4 C-7 a^2 b^2 (3 A+C)-b^4 (7 A+5 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 b^5 d}-\frac {2 a^3 \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 (a+b) d}+\frac {2 \left (7 A b^2-7 a b B+7 a^2 C+5 b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 b^3 d}+\frac {2 (b B-a C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 d}+\frac {2 C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 b d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 2.70, size = 335, normalized size = 1.18 \[ \frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (70 a^2 C+42 b (b B-a C) \cos (c+d x)-70 a b B+70 A b^2+15 b^2 C \cos (2 (c+d x))+65 b^2 C\right )+\frac {4 \left (-28 a^2 C+28 a b B+35 A b^2+25 b^2 C\right ) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{a+b}-\frac {2 \left (35 a^3 C-35 a^2 b B+a b^2 (35 A+13 C)-63 b^3 B\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}-\frac {42 \sin (c+d x) \left (5 a^3 C-5 a^2 b B+a b^2 (5 A+3 C)-3 b^3 B\right ) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a b^2 \sqrt {\sin ^2(c+d x)}}}{210 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x]),x]

[Out]

((-2*(-35*a^2*b*B - 63*b^3*B + 35*a^3*C + a*b^2*(35*A + 13*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a +
b) + (4*(35*A*b^2 + 28*a*b*B - 28*a^2*C + 25*b^2*C)*((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(
a + b), (c + d*x)/2, 2]))/(a + b) + 2*Sqrt[Cos[c + d*x]]*(70*A*b^2 - 70*a*b*B + 70*a^2*C + 65*b^2*C + 42*b*(b*
B - a*C)*Cos[c + d*x] + 15*b^2*C*Cos[2*(c + d*x)])*Sin[c + d*x] - (42*(-5*a^2*b*B - 3*b^3*B + 5*a^3*C + a*b^2*
(5*A + 3*C))*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x
]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b^2*Sqrt[Sin[c
+ d*x]^2]))/(210*b^3*d)

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{b \cos \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^(5/2)/(b*cos(d*x + c) + a), x)

________________________________________________________________________________________

maple [B]  time = 7.88, size = 1097, normalized size = 3.85 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(8/105*C/b*(60*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^8-258*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+448*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+85*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-168*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-167*sin(1/2*d*x+1/2*c)^2*cos(1/
2*d*x+1/2*c))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+4/5/b^2*(B*b-C*a-4*C*b)*(-4*cos(1/2*d*x+1/2
*c)*sin(1/2*d*x+1/2*c)^6+14*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*
d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*
c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-6*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))/(-2*sin(1/2*d*x
+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+4/3/b^3*(A*b^2-B*a*b-3*B*b^2+C*a^2+3*C*a*b+6*C*b^2)*(2*sin(1/2*d*x+1/2*c
)^4*cos(1/2*d*x+1/2*c)+2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1
/2*c),2^(1/2))-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^
(1/2))-sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-2/b^4*(A*
a*b^2+2*A*b^3-B*a^2*b-2*B*a*b^2-3*B*b^3+C*a^3+2*C*a^2*b+3*C*a*b^2+4*C*b^3)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*co
s(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c)
,2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*(A*a^2*b^2+A*a*b^3+A*b^4-B*a^3*b-B*a^2*b^2-B*a*b^3-B*b^4+C*
a^4+C*a^3*b+C*a^2*b^2+C*a*b^3+C*b^4)/b^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*si
n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+4*a^3*(A*b^2-B*a*b+C*a^2)
/b^4/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x
+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{b \cos \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^(5/2)/(b*cos(d*x + c) + a), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{a+b\,\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*cos(c + d*x)),x)

[Out]

int((cos(c + d*x)^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*cos(c + d*x)), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+b*cos(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________